Design of Shallow Foundation (Isolated Footings)

AIM:

To design a square footing for a column size of 400x400 for the given data and to do 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

Introduction:

• Footings and other foundation units transfer loads from the structure to the soil or rock supporting the structure.
• The distribution of soil pressure under a footing is a function of the type of soil and the relative rigidity of the soil and the foundation pad.
• Most common types of footings used are:
  • Strip or wall footing
  • Spread footing
  • Stepped footing
  • Tapered footing
  • Pile cap
  • Combined footing
  • Mat or raft footing

Given data

• Column size = 400x400mm
• Compression axial load (1.5 DL+1.5 LL) = 2000 KN
• Safe bearing capacity of soil = 150 KN/M2 at 2m depth
• Grade of concrete = M25

Calculation of gross soil pressure & footing area:

• Unfactored load with surcharge + self weight of 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.

Net soil pressure & Initial footing depth:

• Factored load / Actual area of footing

= 2000x1000 / (3.2x3.2)

Pn = 195.3 KN/m2

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

√(footing base bearing area/loaded area) ≤ 2

• Permissible bearing stress at Jn = 0.45x Fck = 0.45x25 = 11.25 N/M²

Permissible bearing stress x area factor = 11.25x2 = 22.50 N/M²

Actual bearing stress at base of column

= (2000x1000)/(400x400)

= 12.50 N/M²

Bearing stress at base, 12.50 < 22.50 N/M²

Hence Ok with this depth 500mm.

Taking effective cover as 50mm

Ie effective depth = 450mm

3. Flexural reinforcement calculation :

• Conversion of net pressure into udl

= 195.30 x 3.20 = 625 KN/M

• BM due to udl over cantilever section

= wl2/2 = 625x1.6x1.6/2 (1 = 3.20/2

= 800 KN-m

• Reinforcement calculation as per IS :456 :200 , Annex – G P-96

M = 800 KN-m

Fy = 500 N/mm2

b = 3.20 m

d = 450 mm

fck = 25 N/mm2

Ast = ?

800 = 0.87 x 500 xAstx450 ((1- Astx500/(3200x450x25))

= 195750 Ast – 2.72 Ast

Ast = 4350 mm2

Taking 16mm bar = 4350/201 = 22

16mm bar @ 130mmc/c

Ie 24 x 201 = 4824 mm2, Hence OK

4.One way shear check:

Shear force in the one way shaded area

= 0.95 x 3.20 x 195.30 = 593.70 KN

Design shear stress at one way cut face,

= 593.70x1000 / (3200x450) = 0.412 N/mm2

Shear stress capacity as per IS 456 Table 19

5.Two way shear check:

• Shear force in the two way shaded area

= (3.2x3.2) – (0.85x0.85)x 195.30 = 1858.80 KN

Design shear stress at one way cut face,

= 1858.80x1000 / (3400x450) = 1.215 N/mm2

Shear stress capacity as per IS 456 clause 31.6.3.1

= 0.25 x 5 = 1.25 N/mm2

Design shear (1.215)< Shear stress capacity (1.25),

Hence OK