Design of Shallow Foundation (Isolated Footings)
1) Aim and Introduction: Design of RCC Isolated footing with given data as per IS:456.
2) Procedure in detail:
a) Gross soil pressure & Footing area:
(2000/1.5*1.1) = 9.78 m2
Assuming we have square footing
Side of the footing= side of the square= sqrt(9.78) = 3.12m
Let side of the footing = 3.2m
b) Net soil pressure and initial footing depth:
Factored load/ actual area of footing = 2000*1000/(3.2*3.2)
Pu = 195.3 kN/m2
b)Assume depth of footing 500mm and checknig it for bearing stress as per IS-456-2000 caluse 34.4
sqrt(footing base bearing area/loaded area)<=2
c) Permissible bearing stress 0.45 x Fck = 11.25 N/mm2
Permissible bearing stress multiplied with area factor 11.25x 2 = 22.5 N/mm2
d) Actual bearing stress at column base (2000x1000)/(400x400) = 12.5 N/mm2
12.5 N/mm2 actual stress is less than permissible 22.5 N/mm2. Hence okay, we can proceed with 500mm depth footing for further design checks . If we assume 50mm cover for footing, the effective depth (d) shall be taken as 450mm.
Flexural Reinforcement:
Conversion of net pressure into Uniformly distributed load = 195.3x3.2= 625 kN/m
Bending moment due to UDL over a cantilever section = wL^2/2 = 625x1.6*1.6/2 = 800 kN.m
Reinforcement calculation as per IS-456-2000 Annex-G
Mu=0.87*fy*Ast*d*(1-(Ast*fy/bd/fck))
Ast = 4350 mm2
If we provide 16mm dia. rebar at 125mm spacing, the actual rebar provided = 3200/125*3201
As provided = 5146mm2
One way shear stress check:
Shear force in the one way at a distance d from the face of the column = 0.95x3.2x195.3 = 593.7 kN
Shear stress capacity of sectiona s per IS-456-2000. Table-19= 0.425 N/mm2
Design shear stress at One way out face = 93.7x1000/(3200x450)=0.412 N/mm2
Design shear stress is less than shear capacity
Hence Okay.
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 = 1859*1000/(3400x450) = 1.215 N/mm2
Shear stress capacity of section as per IS-456-2000 clause 31.6.3.1 = 0.25sqrt(25)=1.25 N/mm2
Design shear stress is less than shear capacity
Hence okay.