Design of Shallow Foundation (Isolated Footings)

AIM: To design a square footing for a column size of 400x400 for given constrains and do the following check:

• 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

Also to provide a sketch showing the details of geometric and stuctural features of footing.

INTRODUCTION:

• Footing is a part of foundation which is constructed with concrete or brickwork masonry and acts as a base to the floor columns and floor walls.
• The main function of footing is to transfer the vertical loads directly to the soil.
• The four main types of footing are strip/wall footing,spread footing,combined footing and raft footing.
• A wall footing or strip footing is a continuous strip of concrete that serves to spread the weight of a load-bearing wall across an area of soil.
• Spread footing is a pad which distrinute the colun load in 2 directions to an area of soil around the column.It can be stepped,tappered etc.,
• Combined footing is provided when two columns are closely placed or when column is at property line.
• Raft footing is provded for very weak soil or when area of overall column to the area of site footprint exceeds 50%.
• Isolated footing is individual footing which transfers load to the underground soil. It is provided when a single column is to be provided. 

PROCEDURE/CALCULATION:

 step 1- given data

• column size 400mmx400mm
• compression axial load for the load combination of (1.5 DL + 1.5 LL) is 2000 KN
• safe soil bearing capacity at a depth of 2 metres below E.G.L is 150 KN/m2
• assuming the self weight of the footing and soil above it as 10% load
• grade of concrete is 25kN/m^2
• assuming fy=415 N/mm^2
• here a sloped footing is designed.

step 2- size of footing

• unfactored load= 2000/1.5=1333.33kN
• total unfactored load=1333.33⋅1.1=</math>">1333.33⋅1.1=1333.33⋅1.1=1466.6kN
• area of footing=1466.6150=9.7m</math>">1466.6150=9.7m1466.6150=9.7�
• since its a square footing,B=9.7</math>">B=√9.7�=9.7
• B=L=3.5m,area of footing=12.25 m^2

step 3- net upward pressure

• net upward pressure= factored load*1.5/size of footing
• net upward pressure=2000/12.25=163.26 kN/m^2

step 4-depth of the footing

maximum bending moment at the face of column

length of the critical section=3.5-0.2=1.55m

load on critical section= net upward pressure*length=163.36*=572kN/m

moment=wl^2/2=572*1.55^2/2=683.4kN/m

ultimate moment,Mu= 1.5*686.4=1030kN/m

depth of the footing,D=Mu=0.138*fck*B*d^2

1030⋅106=0.138⋅25⋅3500⋅d2</math>">1030⋅106=0.138⋅25⋅3500⋅d21030⋅106=0.138⋅25⋅3500⋅�2

d=85300=292mm</math>">√85300=292mm85300=292��

assuming thumb rule,d=2.5*292=730mm

as∑∈g50mmcovergiven,d=730+50=780mm≈,800mm</math>">as∑∈g50mmcovergiven,d=730+50=780mm≈,800mm��∑∈�50������������,�=730+50=780��≈,800��

step 5-check for gross bearing capcity

assuming unit weight of the soil is 18kN/m^3

load on the footing=1333.3kN

weightofthef∞t∈g=3.5⋅3.5⋅25⋅0.8=245kN</math>">weightofthef∞t∈g=3.5⋅3.5⋅25⋅0.8=245kN����ℎ����ℎ��∞�∈�=3.5⋅3.5⋅25⋅0.8=245��

weight of the soil=3.5⋅3.5⋅(2-0.8)⋅18=264.6kN</math>">3.5⋅3.5⋅(2−0.8)⋅18=264.6kN3.5⋅3.5⋅(2-0.8)⋅18=264.6��

total load = 1852kN

gross bearing capacity= 1852/3.5*3.5=257 kN/m^2>150kN/m^2,hence we need to increase the size of footing to 3.8mx3.8m

now total load=1934kN

gross bearing capacity= 1934/3.8*3.8=133.9 kN/m^2<150kN>

socorrected≠tsoilpressure=20003.8⋅2.8=138.5kN���or������≠�������������=20003.8⋅2.8=138.5��

step 6-check for bearing pressure at column-footing joint

from IS 456,clause 34.4

`bearing pressure=0.45*fck*(A1/A2)^1/2

bearing pressure=0.45*25*400*400=1800 kN<2000>

step 7-reinforecement

`Pt=50fck/fy*1-1-4.6⋅Mfck⋅B⋅d2</math>">1−√1−4.6⋅Mfck⋅B⋅d21-1-4.6⋅����⋅�⋅�2

t=50⋅25415(1-1-4.6⋅1030⋅10625⋅3800⋅800⋅800=0.126%</math>">t=50⋅25415⎛⎝1−√1−4.6⋅1030⋅10625⋅3800⋅800⋅800=0.126%�=50⋅25415(1-1-4.6⋅1030⋅10625⋅3800⋅800⋅800=0.126%

Ast=Pt*b*d/100=3830mm^2

number of bars= Ast/area of one bar(16mm bar)

number of bar=3830/3.14*16*16/4=20

16*3.14*16*16/4=4019mm^2>3830mm^2

spacing= 201*3800/4019=190mm

lets provide 20# 16mm dia @150mm c/c

step 8-check for one way shear 

the critical section for shear is at distance d from the face of column,

shear force=net soil pressure*((B-b)/2)-d=138.5*((3.8-0.4)/2)*0.8=473.67kN

100 Ast/Bd=100*4020/3800*800=0.132

Tv=Vu/Bd=473.67*10^3/3800*800=0.16

Tc from table 19 of IS 456 is 0.29

Tv

step 9-check for two way shear 

the crititcal section for shear is at distance d/2 from face of column

shear force,Vu= net soil pressure*(B^2-b'^2)=138.5*(3.8^2-1.2^2)=1800.5kN

periphery of section,Bo=2(1.2+1.2)=4.8m

Tc=Vu/Bod=1800.5*10^3/4800*800=0.46 N/mm^2

permissible shear stress=ks*Tc

ks=(0.5=Bc)>1,Bc=400/400=1

permissible shear stress=1+0.5=1.5 N/mm^2.

Tc=0.25fck</math>">√fck���=0.2525=1.25</math>">√25=1.2525=1.25N/mm^2

Tv

step 10- development length

Ld=0.87⋅fs⋅ϕ4⋅τbd</math>">0.87⋅fs⋅ϕ4⋅τbd0.87⋅��⋅�4⋅���

0.87*25*16/4*1.6*1.4=677mm

length available=1600-50=1550mm >677mm

RESULT-

• Sloped isolated footing for square column is desiged for given condition and also checked