Design of Shallow Foundation (Isolated Footings)

ANSWER

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

1. Column load 2000KN
2. 10% of footing self-weight = 2000x10/100 = 200KN
3. Axial load on footing = 2000+ 200 = 2200KN

Area of footing

1. Af = Load on footing / safe bearing capacity = 2200/150 = 14.66m2
2. B = L = Ѵ14.66 = 3.83m Say B = L = 3.90m = 3900mm

Net Upward pressure

1. 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

1. At critical section distance from face of column to edge of footing = l = (3.9 -0.4) /2 = 1.75m
2. Load on critical section = 4 x 216.96 = 867.84KN/m
3. BM=wl^2/2=86784x175x175/2=132888KNm
4. BM w l 2 /2 867.84 x 1.75 x 1.75 / 2= 1328.88KNm

Depth of footing

1. Mu= 0.133 fck b d2 for Fe 500
2. d = Ѵ (Mu / (0.133 fck b))

= Ѵ 1328.88x10^6 / 0.133 x 25 x 4000

= 315.99mm2 Say 650mm

Cover = 50mm

3. Overall Depth of footing = d + C + (bar dia (say 16mm)/2) = 650 + 50 + 16/2 = 708mm Say 750mm

D = 750mm

Reinforcement Requirement

1. 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
2. Ast = 0.167 x 4000 x 700 /100 = 4676 mm2
3. 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

1. 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

2. Tv = V/bd = 238.66/4000x700 = 0.085n/mm2
3. Tc = 0.30 N/mm2 as per table 19
4. Tv < Tc hence shear stress are less than the permissible stress limit

Check for Two-way shear

1. 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

2. Tv = Vu / b0 d = 3155.8 / 4400 x 700 = 1754.44 KN = 1.025N/mm2
3. The permissible shear stress = ks Tc

ks =(0.5+Bc) > 1

Bc= b/b = 400/400 = 1.00

ks = 0.5+ 1 = 1.50

4. Permissible shear stress (Tcxk) (N/mm2) = 1.50
5. 25 x SQRT (fck) = 1.25
6. Tv < Tc
7. 25< 1.50

Hence shear stresses are within safe permissible limits