Design of Shallow Foundation (Isolated Footings)

AIM : To design shallow foundation (Isolated footing)

INTRODUCTION : Footings transfer loads from the structure to the soil or rock supporing structure. Area of footing is based on the allowable stresses acting on the soil at unfactored or service loads.

Given : column size of 400x400

Compression axial load = 2000KN

Soil Bearing capacity : 150KN/m2 at depth 2m

Concrete grade M25 and Steel HYSD 415

PROCEDURE :

• Area required for footing = load / soil bearing capacity
  
  = ((2000x1.15)/1.5)/ 150 (15% backfill soil)
  
  = 10.22 m2
  
  Assume 3.6 x 3.6 m2 size footing
  
  
• Factored soil pressure = 2000/(3.6 x 3.6)
  
  = 0.154 N/mm2
  
  Assume shear strength of concrete is 0.36 according to IS 456 Table 19, ( for M25)
  
  
• Vu = 0.36(3600)d
  
  Vu(actual) = 0.163(3600)(1600-d)
  
  Vu >= Vu(actual)
  
  1296d >= 938880 - 586.8d
  
  d >= 498.66mm
  
  Therefore assume d=500 mm
  
  
• Total depth D = 500 + 50mm cover + 16mm + 8mm (diameter of reinforcing bars both the ways)
  
  = 574mm
  
  
• Assume unit weight of concrete 24KN/m3 and weight of soil 20KN/m3
  
  Service load = 2000/ 1.5 = 1333.33 KN
  
  Weight of footing = 3.6 x 3.6 x 0.574 x 24 = 178.53 KN
  
  Weight of soil = 3.6 x 3.6 x (2-0.574) x 20 = 369.61 KN
  
  Total load = 1333.33 + 178.53 + 369.61 = 1881.47 KN
  
  Gross bearing pressure = (1881.47/3.6x3.6) = 145.175 KN/mm2 < 150KN/m2 Hence ok
  
  
  
• Bending moment
  
  The critical section is at the face of column
  
  Mu = 0.154 x (3600) x (1600) x (800) = 709.632 KNm
  
  Mu = 0.87FyAstD (1-(AstFy/FckBD))
  
  709.632 x 10^6 = 0.87 x 415 x Ast x 574 (1- (Ast 415/25x 3600x 574)
  
  Ast = 3523.61 mm2
  
  Assuming p=0.25%
  
  Ast = 0.0025 x 3600 x 574 = 5166 mm2
  
  26 bars of 16mm diameter (5227mm2) both ways
  
  The spacing of bars = (3600 - 2(50) - 16)/17 = 204.9mm = 200mm c/c distance
  
  
  
• Check for 2 wayshear
  
  Critical section is set at d/2 from periphery of column
  
  The factored shear force = 0.154((3600^2 - (400 +500)^2 x 10^-3 = 1871.1 KN
  
  Shear strength = Ks `tau`c = Ks 0.25 x Fck ^(1/2)
  
  `beta` = 1, Ks =0.5 +`beta`
  
  Ks = 1.5 since greater than one, Ks = 1
  
  therefore Shear strength = 0.25 Fck^(1/2)
  
  = 1.25 N/mm2
  
  Shear resistance = 1.25 x 4 x 900 x 500 = 2250 KN > 1871.1 KN
  Hence ok
  
  Depth of footing govered by one way shear
  
  
• Longitudinal reinforcement
  
  Asssue 12 bars of 20mm
  
  
• Shear reinforcement
  
  Assume 8mm bars at 300c/c
  
  
• Development length (According to IS 456 clause 26.2.1)
  
  Ld = Fs `phi` / 4 `tau`
  
  = (0.87 x 415 x 16 )/ 4 x 1.6 x 1.4
  
  = 644.73 mm
  
  Length available = 1600 - 50 = 1550 > 644.73 mm
  
  Pu = 2000KN
  
  Compressive bearing resistance = 0.45 Fck `((A1)/(A2) )^ (1/2)`
  
  = 0.45 X 25 X 400 X 400
  
  = 1800 KN < 2000KN
  
  200KN to be transferred by longitudinal reinforcement
  
  12 bars, 20mm diameter bars assumed for footing, stress level = 0.87Fy(200/2000) = 36.105
  
  Required length = 36.105 x 25 / 4 x 1.6 x 1.25 x 1.4 = 80.59
  
  Length available = 574 - 50 - 16 - 16 = 492mm > 80.59mm
  
  Hence ok
  
  
  Result :