Design of Shallow Foundation (Isolated Footings)
Column Size = 400 x400
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 = 2metres
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 = Ѵ14.66 = 3.83m Say B = L = 3.90m = 3900mm
Net Upward pressure
Factored net soil pressure = load on footing /area of footing
= 2200 / ( 3.90 x 3.90)
= 144.64 x 1.5 (factor)
= 216.96KN/m2
- Bending Moment
At critical section distance from face of column to edge of footing = l = (3.9 -0.4) /2 = 1.75m
Load on critical section = 4 x 216.96 = 867.84KN/m
BM = w l ^2 /2 = 867.84 x 1.75 x 1.75 / 2 =1328.88KNm
- Depth of footing
Mu= 0.133 fck b d2 for Fe 500
d = Ѵ ( Mu /( 0.133 fck b) )
= Ѵ 1328.88x10^6 / 0.133 x 25 x 4000
= 315.99mm2 Say 650mm
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/bd2 = 1328.88 x10^6 / 4000 x 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 mm2
Numbers of bar = 4676/ 201 = 23.26 Say 24Nos
24X201 = 4824mm2 > 4676mm2 Spacing = 201 x 4000/ 4824 = 166.66mm 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 = Upward soil pressure x (B - b) / 2 -d = 216.96 x (4-0.4)/2 -0.7 = 238.66KN
100 Ast / b d = 100 X 4824 / 4000 x 700 = 0.172
Tv = V/bd = 238.66/4000x700 = 0.085n/mm2
Tc = 0.30 N/mm2 as per table 19
Tv < Tc hence shear stress are less then the permissible stress limit
- Check for Two way shear
The critical section for two way shear is located at a distance of d/2 from the face of the column
Shear force on critical section
V = 238.66 ( 4x4 – 1.1 x 1.1) =3155.8 KN
Periphery of the section b0 = 2 ( 1.1 + 1.1) = 4.4m = 4400mm
Tv = Vu / b0 d = 3155.8 / 4400 x 700 = 1754.44 KN = 1.025N/mm2
The permissible shear stress = ks Tc
ks =(0.5+Bc) > 1
Bc= b/b = 400/400 = 1.00
ks = 0.5+ 1 = 1.50
Permissible shesr stress (Tcxk) (N/mm2) = 1.50
0.25 x SQRT (fck) = 1.25
Tv < Tc
1.25 1.50
Hence shear stresses are within safe permissible limits