Design of Shallow Foundation (Isolated Footings)

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).

Given data:

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.  

Steps by step procedure of 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.  

Calculation:

1. Calculation of gross soil pressure for footing area:

Unfactored load with surcharge+ self-weight of the footing/ allowable soil pressure.

 q_a is allowable soil pressure.

Assume adding 10 % of surcharge and self-weight as it is not given in the data.

A = (2000/1.5 X1.1) KN/ 150 KN/m² = 9.78 m² 

Assume it as square footing.

A= √9.78 = 3.12-meter side footing.

Assume, 3.2 meter sided isolated footing

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

Net soil pressure is excluding self-weight and soil surcharge.

P_n=Factored load/ Actual area of footing.

=2000/ (3.2 x3.2)

P_n = 195.3 KN/m²

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

√2400/400 =2.44, which is greater than 2.

So take it as  2.

 

Permissible bearing stress 0.45 x Fck = 0.45 x 25 = 11.25 N/mm².

Permissible bearing stress multiplied with area factor = 11.25 x2 = 22.5 N/mm².

Actual bearing stress at column base= (2000 x 1000) / ( 400 x 400)

= 12.5 N/mm².

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

The design can be carried on with 500mm depth footing for further designing process.

If we assume 50mm cover for footing,

the effective depth(d) = over all depth – cover = 500-50 =450mm

d= 450mm ; D=500mm

3.Flexural reinforcement:

P_n = 195.3 KN/m²

Conversion of net pressure into uniformly distributed load = 195.3 x 3.2

= 625 KN/m (Converting KN/m2 in to KN/m for UDL)

Bending moment due to UDL over a cantilever section = wL² / 2

Over all length of the footing is 3.2, so half the length ,L= 1.6

= 625 x 1.6 x 1.6 / 2.

M= 800 KNm.

Reinforcement calculation as per IS-456 –  2000 Annex G

800= 0.87 x500 x Ast x 450( 1 –((Ast x 500)/(3.2 x 450 x 500))

Assume Fy = 500 N/ mm2,

 In result Ast required = 4350 mm2.

If we provide 16mm dia rebar @ 125mm spacing, the actual rebar provided

 =3200/ 125 x 201

Ast provided = 5146 mm2.

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 T_v=593.7 x 1000 /(3200 x 450)

 =0.412 N/ mm2

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

=0.64 N/mm2

Design shear stress is less than shear capacity. Hence ok.

5.Two-way shear stress check:

d/2= 450/2 =225, That’s where crack occurs for 2 way shear.  

Shear force in the two way shaded area=  ( 3.2 x 3.2 ) – ( 0.85 x 0.85) x 195.3

 = 1859KN.

The over all length of the cut session = 850 +850 +850 +850=3400, d =450

Design shear stress at Two way cut face T_v= 1859 x1000 / (3400 x 450 )

=1.215 N/ mm2

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

= 0.25 X √25 = 1.25 N/mm2.

Design shear stress is less than shear capacity.

Hence ok.

Result:

Using uniform depth session for footing design as in one way shear and 2-way shear check we used d =450 and D= 500.

Here, using 16mm diameter bar in both direction with 125mm spacing.

Development length = 41 times the diameter of the bars =41 x 16 =656mm