Design of Shallow Foundation (Isolated Footings)

CHALLENGE 11:

Design of Shallow Foundation (Isolated Footings)

AIM:

To design of Shallow Foundation (Isolated Footings)

INTRODUCTION:

The isolated footing is the simplest type of foundation. It is a single column of concrete or masonry that transfers a load of a structure to the soil. The footing must be large enough to distribute the load over a sufficient area of soil.

QUESTION:

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.

Please do go through the solved examples in the document provided with the challenge.

PROCEDURE:

Steps involved in Footing design:

1. Calculation of gross soil pressure for footing area

2. Calculation of Net soil pressure for flexure and shear design

3. Initial depth of footing calculation

4. Flexural reinforcement calculation

5. One-way shear check

6. Two-way shear check

STEP-1:

Calculation of gross soil pressure for footing area:

=(unfactored load with surcharge + self-weight footing )/ allowable soil pressure

=(2000/1.5 x 1.1) / 150= 9.78 m2
Assuming we have square footing

√9.78 = 3.12-meter side footing

Consider, 3.2 meter-sided isolated footing

STEP-2 &3:

Calculation of Net soil pressure for flexure and shear design & Initial depth of footing calculation

• Factored load / actual area of the footing

2000x1000 / (3.2x3.2)

Pn=195.3 kN/m2

Assume the depth of footing is 500 mm and check it for bearing stress as per IS-456-2000 clause 34.4

√(footing base bearing area / loaded area) <=2

• Permissible bearing stress 0.45 x Fck= 11.25 N/mm2

Permissible bearing stress multiplied with area factor 11.25 x 2=22.5 N/mm2

• Actual bearing stress at the column base

(2000x1000) / (400x400) = 12.5 N/mm2

12.5 N/mm2 actual stress is less than perrissible 22.5 N/mm2. Hence Okay.

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

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

STEP-4:

Flexural reinforcement calculation:

Conversion of net pressure into Uniformly distributed load =195.3 x 3.2= 625 kN/m

Bending moment due to UDL over a cantilever section

= (wL^2)/2=625x1.6x1.6/2

M= 800 Knm

• Reinforcement calculation as per IS-456-2000 Annex-G

Assume Fy=500 N/mm2, In result Ast required = 4350 mm2

If we provide 16mm dia rebar at 125 mm spacing, the actual rebar provided = 3200/125 x 201 Ast provided = 5146 mm2

STEP-5:

One-way shear check:

• Shear force in the one-way shaded area

=0.95 x 3.2 x 195.3=593.7 kN

• Shear stress capacity of section as per IS-456-2000 Table-19 (include rebar)

=0.425 N/mm2

• Design shear stress at way cut face

Tv =593.7x1000 / (3200x450) = 0.412 N/mm2

Design shear stress is less than shear capacity.

Hence Okay

STEP-6:

Two-way shear check

• Shear force in the two-way shaded area

=(3.2x3.2) - (0.85x0.85) x 195.3=1859 kN

• Shear stress capacity of section as per IS-456-2000 clause 31.6.3.1

=0.25 x JFck= 1.25 N/mm2

• Design shear stress at Two way cut face

Tv =1859x1000 / (3400x450) =1.215 N/mm2

• Design shear stress is less than shear capacity.

Hence Okay

Reinforcement Details:

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