Design of Shallow Foundation (Isolated Footings)

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

OBJECTIVE OF THE CHALLENGE:

• To know the General design procedure of Spread footing with hand calculations
• To practice how the inputs can be collaborated and arrived as data for calculations (How assumptions are helpful when certain input is not available).

DATA GIVEN:

Size of the Column = 400X400

Compression axial load = 2000 KN

Safe Soil Bearing Capacity = 150 KN/m2

Grade of concrete = M25

We have to Consider square footing as given.

PROCEDURE:

1. Calculations of gross soil pressure for footing area:

Unfactored load with surcharged + self weight of footings/ allowable soil pressure

= (2000/1.5 X 1.1) / 150 = 9.78m²

Assuming we have square footing

Now calculate meter side footing

√9.78 = 3.12

Consider 3.12 as Meter Side Footing.

2. Net soil pressure for flexure and shear design:

• Net pressure = Factored load /actual area of footing

2000 x 1000 / (3.2 x 3.2)

P_n= 195.3kN/m²

3. Initial depth of footing calculation:

• Assume depth of footing 500mm and checking 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 is:

11.25 x 2 = 22.5 N/mm²

• Actual bearing stress at column base is given by:

(2000X1000)/(400 x400) = 12.5 N/mm²

Where;

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

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

If we assume with 500mm cover for footing, the effective depth (d) shall be taken as 450mm.

4. Flexure reinforcement calculations:

• Conversion of net pressure into uniformly distrusted load

= 195.3 x3.2 = 625kN/m

• Bending moment due to UDL over a cantilever section

wL²/2 = 625x1.6x1.6/2 = 800kNm

i.e. M = 800kNm

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

Assume Fy=500 N/mm²,

In result Ast required = 4350mm²

If we provide 16mm dia rebar at 125mm spacing, the actual rebar provided is given by

= 3200/125 x 201

Ast provided = 5146mm²

5. 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

(ꚍ_v)= 593.7x1000/(3200x450) = 0.412 N/`mm^2`

• Shear stress capacity of section as per IS-456-2000 Table-19 (include rebar) = 425 N/`mm^2`
• Design shear stress is less than shear capacity. Hence Okay.

6. 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

(ꚍ_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:

We have succesfully understand and design the isolated footing as given above using Auto CAD.