Design of Shallow Foundation (Isolated Footings)

AIM

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 meter 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

INTRODUCTION

Here we are designing a square footing for a column of 400x400 size

The reinforcement detail drawing is drawn using AutoCAD

Code used 1S 456-2000

PROCEDURE

Step 1:Calculation of the size of the footing

Load acting on the footing=unfactored load on column + self-weight of footing/allowable soil pressure

=(2000/1.5)+2000*0.1=1533 KN

Area of footing=(Load acting on the footing/safe bearing capacity)

=1533/150 =10.22 mm^2

Assume square footing

Area of footing =A^2

A=square root of 10.22

A=3.19

Length=width =3.20m

Step 2:Calculation of Net soil Pressure

Net soil Pressure=Factored load/Area of footing

=2000/(3.2*3.2)

=195.3 KN/m^2

Step 3:Determination of initial depth of footing

Assuming the depth of footing to be 500 mm and checking it for bearing stress as per IS 456:2000 clause 34.4

As per the IS 456:2000 clause 34.4,

√(footing base bearing area/loaded area at the column base)≤2

=√(2400X2400)/(400X400)=6 ,as it is greater than 2 ,we cannot use the value

Permissible bearing stress=0.45XFck=0.45X25=11.25N/mm^2

Permissible bearing stress multiplied by area factor=11.25X2=22.50N/mm^2

The actual bearing stress at the column base=(2000X1000/400X400)=12.5N/mm^2

Actual bearing stress is lesser than the permissible bearing stress,12.5<22.5N/mmm^2, Hence,ok.

We can proceed with 500mm depth for further design checks.

If we consider 50 mm as cover,then the effective depth can be =d=450mm.

Step -4:Determination of Flexural Reinforcement

Conversion of net pressure into uniformly distributed load=195.3*3.2=624.96

=625KN/m

Bending moment due to udl over a cantilever section=(W*L^2)/2=(625 X 1.6*1.6)/2

=800 KNm

Reinforcement calculation as per IS 456:2000, Annex G

Assume fy = 500N/mm^2

Then the Ast =4350mm^2

Provide 16 mm dia bars at 125 mm c/c spacing

Number of bars=3200/125=25.6

Provide 26 bar

Ast provided =26*201=5226

Step -5:Check for One Way Shear

Critical setion d from face of column

Shear force in the one way Shear shaded area=195.3 X 0.95 X 3.2=593.7KN

Design shear stress at one way cut face,

τv = Vu/Bd=593.7x1000/3200X450=0.412 N/mm^2

Shear stress capacity of the section as per IS 456:2000,Table 19

We arrived the Tc value as =0.425 N/mm2

Since,the design shear stress is less than the shear capacity, hence,ok.

Step -5:Check for Two Way Shear

Critical setion d/2 from face of column

Shear force in the Two way Shear shaded area=(3.2X3.2)-(0.85x0.85)x195.3=1859 KN

Design shear stress at Two way cut face,

τv = Vu/Bd=1859X1000/3400x450=1.215 N/mm^2

Shear stress capacity of section as per is 456:2000 clause 31.6.3.1

0.25√Fck=0.25*25=1.25 N/mm^2

The design shear stress is less than the shear capacity, hence,ok

RESULTS