Design of Shallow Foundation (Isolated Footings)

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

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:

• Gross soil pressure = (Unfactored load + surcharge + self weight of footing) / safe bearing capacity
  
  = 20001.5⋅1.1150${\displaystyle \frac{\frac{2000}{1.5}\cdot 1.1}{150}}$ (Adding 10% for surcharge and self weight)
  
  = 9.78 m2
  
  
• Assume square footing
  √9.78 = 3.12 m
  Provide 3.2 x 3.2 m footing size

Calculation of net soil pressure & initial footing depth

• Net soil pressure = Factored load/actual area of footing
  = 20003.2⋅3.2${\displaystyle \frac{2000}{3.2\cdot 3.2}}$
  = 195.30 kN/m2
  Assume initial overall depth of footing as 500mm and checking for bearing stress as per IS 456-200 clause 34.4
  
  
  
  

  Permissible bearing stress at column footing joint = 0.45x Fck = 0.45x25 = 11.25 N/m2
  Permissible bearing stress x area factor = 11.25 x 2 = 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 500 mm.

  Taking effective cover as 50 mm

  le effective depth = 450 mm

  
  

Calculation of Flexural reinforcement

• Conversion of net pressure into udl = 195.30 x 3.20 = 625 kN/m

  Bending moment due to udl over cantilever section
  = wl2/2 = 625⋅1.622${\displaystyle 625\cdot \frac{{1.6}^2}{2}}$
  = 800 kN-m

  Reinforcement calculation as per IS :456 :200, Annex G
  
  

  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 x Ast x 450 ((1- Astx500/(3200x450x25))
  = 195750 Ast – 2.72 Ast
  
  Ast = 4350 mm2
  
  Taking 16mm bar = 4350/201.06 = 22
  Spacing = 3200/22 = 145 mm
  Provide 16mm bar @ 125 mm c/c
  The actoal ast provided = 5147 mm2
  Hence OK
  
  
  

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
  
  
  

  For 100Ast/bd = 0.335

  Interpolate those values to get for 0.335
  Shear stress capacity = 0.425 N/mm2

  Design shear < Shear stress capacity
  Hence safe

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
  
  
  
  
  

  RESULT

  Using the given details the square footing has been designed successfully.