Design of Shallow Foundation (Isolated Footings)

Design a square footing for a column size of 400x400. 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 a depth of 2 metres below E.G.L. Participants are free to go for either a tapered or stepped footing. Besides the total axial load, also account for the self-weight of the footing and soil above it. Assume M25 concrete grade. Do the following checks:

• 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.

PROCEDURE:-

The compressive axial load = 2000KN

Unfactored load = 2000/1.5 = 1333.3 KN

The safe bearing capacity = 150 KN/m^2

The grade of concrete = M25

Assume the thickness of the concrete footing to be 0.8m

Depth of 2m below E.G.L

Soil depth = 2-0.8 = 1.2m

Soil density = 20KN/cu.m

RCC density = 25 KN/cu.m

Solution:

The safe bearing should be greater than gross soil pressure

150 > (1333.3/b^2)+(1.2 X 20)+(0.8 X 25)

=150 > (1333.3/b^2) + 44

b=3.54 = 3.7

Gross soil pressure=(1333.33/3.7^2)+44

= 141.3.9 KN/sq.m (Hence okay as it is less than 150)

Net soil pressure:

= 2000/13.69 = 146.09kN/sq.m = 0.146N/sq.mm

Calculation of Bearing pressure:

=Net load/column area = 2000 x 10^3/ 400^2 = 12.5 N/sq.m

Allowable Bearing pressure:

Allowable bearing pressure = 2 x 0.45 x 25 = 22.5 N/sq.mm

Hence safe.

Calculation of one-way shear check:

concrete cover = 60mm

The effective depth of footing = 800-60 = 740mm

= 0.2 + 0.74 + X = 3.7/2

X= 0.91

Net load = 910 x 3700 x 0.146 = 491582 N = 491.582 KN

Design footing area = 3.7 x 0.74 = 2.7382 m^2

One way shear stress = 491.582/2.7382 = 179.527 KN/m^2 = 0.179 N/mm^2

Hence safe.

Calculation of two-way shear check:

C1 + d = 0.4 + 0.74 = 1.14 m

C2 + d = 0.4 + 0.74 = 1.14 m

Net load = (0.146 x 3.7 x 3.7) - (0.146 x 1.14 x 1.14)

=1.18089N = 1808.9 KN

Perimeter = 4x 0.74 x 1.14 = 3.3744 m^2

Two way shear stress = 1809.9/3.3744 = 536.065 KN/m^2

according to the clause:

0.4/0.4 =1

Calculation of Flexural strength:

w=0.146 x 1000 =146

l=(3700-400)/2 = 1650

146 x 1650^2/2 = 198742500 Nmm

= 198.74 KN-m

One-way shear reinforcement:

One-way shear stressw= 0.179 N/mm^2

As per table 19 of IS 456:2000

= 1110 sq.mm/meter width

shall provide 2 reinforcement of diameter 16mm with 181 mm spacing

1000/181x 201 = 1110.4 mm^2

Calculation of development length:

Diameter of bar = 16mm

Stress in bound = 0.55 Fy = 0.55 x 500 = 275 N/sq.mm

Design bound stress = 1.4

Footing reinforcement:

RESULT:

Hence design of Isolated footing completed.