Design of Shallow Foundation (Isolated Footings)

AIM:the main aim of this challenge is to do design of shallow foundations and sketch the details of footing

INTRODUCTION:

• in this, i am gonna use the data for design given in the question

i.e column size=400mm*400mm

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

Assume M25 concrete grade, fe500 steel grade.

• during the design, the below points needs to be checked and satisfied.

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

• The following steps are involved in footing design

calculation of gross soil pressure for footing area

calculation of net soil pressure for flexire and shear design

initial depth of footing calculation

flexural reinforcement calculation

one way shear check

two way shear check

PROCEDURE:

• starting with the given data to design footing
• gross soil pressure and footing area

Pgross=(unfactored load with surcharge+self weight of footing)/allowable soil pressure

here, we don't have the surcharge or self weight loads. so, we assume 10% from the given compression axial load

hence, Pgross=((2000*1.1)/1.5)/150=9.78m^2

assuming, we have square footing i.e square root of 9.78 is equal to 3.12 meter side footing

consider, 3.2 meter sided isolated footing

• net soil pressure and initial footing depth

Pnet=factored load /actual area of footing

Pnet=2000/(3.2*3.2)=195.3 kn/m^2

assuming depth of foting 500mm and checking it for bearing stress as per IS 456-2000 clause 34.4

square root of ((2.4*2.4)/(0.4*0.4)) less than or equal to 2

=> 6 is not less than or equal to 2

so, we take 2

• permissible bearing stress= 0.45*Fck=11.25 kn/m^2

permissible bearing stress multiplied with area factor=11.25*2=22.5 kn/m^2

• actual bearing stress at column base=(2000*1000)/(400*400)=12.5 n/mm^2

12.5 n/mm^2 actual stress is less than permissible 22.5 n/mm^2 , hence okay

we can proceed with D=500mm depth footing for furthur design checks,

if we assume 50mm cover for footing, the effective depth shall be taken as d=450mm

• flexural reinforcement:

• conversion of net pressure into uniformly distributed load =195.3*3.2=625 kn/m

• bending moment due to UDL over a cantilever section=(WL^2)/2=(625*1.6*1.6)/2= 800 kn-m
• reinforcement calculation as per IS 456-2000, annex G

if we use 16mm dia bars at 150mm center to center distance, then the Ast provided is less than Ast required. so, we can use the 16mm dia bars at 125mm center to center

then, the actual rebar provided=(3200*201)/125

=> Ast provided= 5146mm^2

• one way shear stress check:

• shear force in the one way shaded area =area * soil pressure=0.95*3.2*195.3=593.7 kn
• design shear stress at one way cut face

• shear stress capacity of section as per IS 456-2000, table-19 = 0.425 n/mm^2 using M25 concrete grade and value of 100As/bd

design shear stress is less than shear capacity, hence okay

two way shear stress check:

• shear forces in the two way shaded area=((3.2*3.2)-(0.85*0.85))*195.3=1859 KN
• design shear stress at two way cut face

=(1859*1000)/(3400*450)=1.215 N/mm^2

• shear stress capacity of section as per IS 456-2000, clause 31.6.3.1

=0.25*square root of Fck=1.25 N/mm^2

design shear stress is less than shear capacity, hence okay

RESULT:

the reinforcement details of footing is as follows