Design of Shallow Foundation (Isolated Footings)

Column size = 400 x 400

The compression axial load for the load combination of (1.5 DL + 1.5 LL) is 2000 KN

The safe soil bearing capacity is 150 KN/m2 at foundation depth below E.G.L = 2 metres

Grade of concrete for footing = M25

Footing size

column load 2000KN

10% of footing self weight + 2000x10/100+ 200KN

Axial load on footing = 2000+ 200 = 2200KN

Area of footing

Af= Load on footing / safe bearing capacity= 2200/150= 14.66m2

B=L= v14.66= 3.83m say B=L = 3.90m = 3900mm

Net Upward pressure

Factored net soil pressure = load on footing/area of footing

= 2200/ (3.90x3.90)

=144.64 x 1.5 (factor)

=216.96KN/m2

Bending moment

at critical section distance from face of column to edge of footing= I = (3.9- 0.4)/ 2= 1.75m

Load on critical section = 4x216.96= 867.84 KN/m

BM= wl^2/2= 867.84 x 1.75 x 1.75/2 = 1328.88 KNm

Depth of footing

Mu= 0.133 fck b d^2 for Fe 500

d= v(Mu/(0.133 fck b))

= v1328.88x10^6 / 0.133 x 25 x 4000

=315.99mm^2 say 650 mm

cover = 50mm

Overall depth of footing = d=c= (bar dia (say 16mm)/2)= 650+50+16/2= 708mm say 750mm

D=750mm

Reinforcement requirement

Mu/bd^2 = 1328.88 x 10^6/ 4000x 700^2 = 0.678 say 0.7 for 0.7 Fy = 500 pt = 0.167 % from Sp- 16 table 3

Ast = 0.167 x 4000 x 700/ 100= 4676 mm^2

Numbers of bar = 4676/201 = 23.26 say 24 no.s

24x201 = 4824mm^2 > 4676mm^2 spacing = 201x4000/4824 = 166.66 mm c/c say 16mm bar @ 150mm c/c

check for one way shear

The critical section for shear at the distance d from the face of the column

shear force = Upwar soil pressure x (B-b)/ 2-d = 216.96 x (4-0.4)/2-0.7= 238.66 KN

100 Ast/ bd= 100x4824 / 4000x700 = 0.172

Tv= V/bd = 238.66/ 4000x 700 = 0.085n/ mm^2

Tc = 0.30 N/mm^2 as per table 19

Tv< Tc hence shear stress are less than the permissible stress limit

Check for two way shear

The critical section for two way shear isloacated at a distance of d/2 from the face of the column

Shear force on critical section

v= 238.66 (4x4 - 1.1x1.1)= 3155.8 KN

Periphery of the section b0= 2 (1.1+1.1)= 4.4m= 4400mm

Tv = Vu / b0d = 3155.8/ 4400x 700 = 1754.44 KN =1.025/ mm^2

The permissible shear stress = ks Tc

ks =(0.5+Bc) >1

Bc=b/b= 400/400= 1.00

ks= 0.5+1=1.5

permissible shear stress (Tcxk) (N/mm^2)= 1.50

0.25x SQRT (fck)= 1.25

Tv < Tc

1.25 1.50

Hence shear stress are within safe permissible limits

Reinforcement detail for footing with drawing as follow below.