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.

AIM:

Manually design a square footing for a column with given criteria and to know the general design procedure of spread footing

INTRODUCTION:

In the given criteria designed an isolated footing and provide sketch or reinforcement detail

GIVEN DATA:

Size of the Column = 400X400 

Compression axial load = 2000 KN

Safe Soil Bearing Capacity = 150 KN/m2

Grade of concrete = M25

Consider square footing

PROCEDURE:

Step 1) Gross Soil Pressure & Footing Area

Assuming the 10% acting load will be surcharge and self-weight

Size of the footing

Unfactored load=`frac{2000}{1.5}` = 1333.33KN

Area of footing =`frac{1333.33xx1.1}{150}` =`9.78m^2`

For square footing b = `sqrt 9.78`

b=3.12m`~~3.2`m

Consider, 3.2X3.2m isolated footing, so that actual area of footing `10.24m^2`

Step 2) Net Soil Pressure & Initial Footing Area

Net upward Pressure (Pn) =  Factored Load /actual area of footing

Pn = 2000/10.24 = 195.3kN/`m^2`

• Assume depth of footing 500mm and checking it for bearing stress as per IS-456-2000 clause 34.4

                                       `sqrt`(Footing base bearing area/loaded area) `<=2`

• Permissible bearing stress 0.45 x Fck = 11.25 N/`mm^2`

                                      Permissible bearing stress multiplied with area factor 11.25 x 2 = 22.5 N/`mm^2`

• Actual bearing stress at column base = (2000X1000)/(400 x400) = 12.5 N/`mm^2`

                                     12.5N/`mm^2` actual stress is less than permissible 22.5N/`mm^2`. Hence ok

We can proceed with 500mm depth footing for further design checks

If we assume 50mm cover for footing, the effective depth (d) shall be taken as 450mm

 

Step 3) Flexural Reinforcement:

• Conversion of net pressure into uniformly distrusted load = 195.3 x3.2 = 625kN/m
• Bending moment due to UDL over a cantilever section(M) `bb "w"L^2`/2 = 625x1.6x1.6/2 = 800kNm
• Reinforcement calculation as per IS-456-2000 annex G

Assume Fy=500 N/`mm^2`, in result Ast required = 4350`mm^2`

If we provide 16mm dia rebar at 125mm spacing, the actual rebar provided = 3200/125 x 201

Ast provided = 5146`mm^2`

Step 4) one way shear stress check

• Shear force in the one-way shaded area = 0.95 x 3.2 x 195.3 = 593.7 kN
• Design shear stress at One way cut face (`tau`v)= 593.7x1000/(3200x450) = 0.412 N/`mm^2`
• Shear stress capacity of section as per IS-456-2000 Table-19 (include rebar) = 0.425 N/`mm^2`
• Design shear stress is less than shear capacity. Hence Okay

Step 4) two-way shear stress check

• Shear force in the two-way shaded area = (3.2x3.2) - (0.85x0.85) x 195.3 = 1859 kN
• Design shear stress at Two way cut face (`tau`v) = 1859x1000/(3400x450) = 1.215 N/`mm^2`
• Shear stress capacity of section as per IS-456-2000 clause 31.6.3.1 = 0.25x`sqrt`Fck = 1.25`mm^2`
• Design shear stress is less than shear capacity. Hence Okay

Result:

hence we successfully design the isolated footing are result provide