Design of Shallow Foundation (Isolated Footings)

If the gross bearing pressure under the footing is within safe bearing capacity

Provide sufficient depth and longitudinal steel for resisting bending and shear

Sufficient anchorage for column steel into the footing

If the bearing pressure at column-footing joint is within 0.45 times the maximum concrete compressive strength

Provide a simple sketch showing the size of the footing, maximum and minimum depth of the section, flexural reinforcement in both directions and anchorage of column steel.

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

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

• Reinforcement detail for footing with drawing as follow below

Please find the attached answer sheet