Design of Shallow Foundation (Isolated Footings)
AIM -
- provide Sufficient anchorage for column steel into the footing
- to provide sufficient depth and longitudinal steel for resisting bending and shear
PROSEDURE -
GIVEN DATA
- Size of footing = 400x400mm
- soil bearing capacity = 150 kn/m^2
- compression axial load = 2000 kn
- concrete grade = M25
STEP 1. SIZE OF FOOTING
Assume self weight of footing = 0.1 x 2000 = 200kn
Total Load = 2000+200 = 2200 Kn
Area of footing = 2200/150 = 14.66 kn/m^2
lete take L=B
So, L= `sqrt 14.66` = 3.82 ~ 4
Take 4x4 m size of square footing with area of 16m^2
STEP 2.
upward pressure = w x 1.5/(area of footing)
= 2200 x 1.5/16 = 206.25 kn/m^2
STEP 3. DEPTH OF FOOTING
max B.M at the face of column is given by M= ( PO*B*(B-b)^2)/8
= (206.25x4x(4-0.4)^2)/8
=1336.5 kn/m
Ultimate B.M = 1.5 M = 1.5x1336.5 = 2004.75kn/m
Depth of footing = 0.138Fck Bd^2
=952.85
D = 952.85 + 50 = 1003mm
STEP 4. AREA OF REINFORCEMENT
Ast = (0.5*fck/fy) x [1- `sqrt(1-(4.6*Mu)/F*ck*b*d^2` ]*Bd
= 5985.3 mm^2
% of reinforcement = 0.157%
Assumed 20mm dia
spacing of bar = 200mm c/c
STEP 5. ONE WAY SHEAR
Shear force = PoB [1/2*(B-b)-d]
=699.6 kn
Ultimate shear force = 1.5 x shear force = 1.5 x 699.6
= 1050 kn
AS PER IS 456:2000
Nominal shear stress = Vu/Bd = 0.27 N/mm^2
Design shear strength of concrete = 0.28 n/mm^2
HENCE OK
Vu/Bd= k.t
From here we got d= 937mm<d provided
STEP 6. TWO WAY SHEAR
Width = 400+953 = 1353mm
Net Shear force acting on perimeter = Po(B^2-b^2)
= 449.10 kN
Ultimate shear force = 1.5 x 449.10 = 673.63kN
NOW FROM IS 456:2000
`tau`=0.23 kN/mm^2
permissible `tau`0.25`sqrtfck`=1.25
HENCE OK