Design of Shallow Foundation (Isolated Footings)

Given-

Column size= 400x400 mm

SBC=150 KN/mm2

Compression axial load=2000kN

Fck=M25

Steps to design a square footing-

1. Size of footing
2. Net upward pressure
3. Bending moment
4. Depth of footing
5. Reinsforcement
6. One way and two way shear

Step 1

Size of footing-

Load on column=2000kN

Add 10% of column axial load for footing=10%(2000)

=2000+200

=2200kN

Area of footing-

Area of footing= load on footing /safe bearing capacity

=2200/150

=14.66m2

Root of 14.66=3.82

=4mx4m

So ,provided area=16m2

Step 2=

Net upword pressure=

ρ=force/area

ρ=1.5x total load on footing/area

=1.5x2200/4 x4

=206.25kN/m2

Step 3=

Bending moment=

At the critical section of the footinh face of the column from the edge of footing

(4000/2)-(400/2)=1800mm

So,load on critical section=upward pressure x breadth of footing

=206.25 x 4

=825 kN/m

So,the bending moment=wl~2/2

=1336.5 kN/m

Step 4-

Depth of footing-

Mu=0.138FcKbd2

=0.138x25x4000xd2

1336.5x10~6=0.138 x25x4000xd2

d=311.20mm

by the thumb rule we can take 2 to 2.5 times higher value of d for shear check

d=778mm

D=d+effective cover=bar dia+half of bar dia

=778+50+20+10

D=860mm

Step 5=

Reoinforcement

Bending moment=1336.5x10~6

Percentage of reinforcement=(50Fck/Fy)(1-√4.6x13336.5x10~6/25x4000x778²)

So, Pt=0.1570%

Area of steel=Pt/100 x bd

=0.157/100x 4000x778

Ast=4885mm²

No of bars-

Take 20mm bar

No of bar=4885/314.5

=15.55=16 no.

Provide 16 no of 20 mm bar

Ast provide =16x314.5

=5032mm²

Here,Ast>Ast provide

4885>5032

Hence ok.

Step 6-

Check for clear spacing between bars

As per is 456 table 15 for Fe415, the clear spacing between the bar should not be less than 180mm

C/C spacing =(4000-50-50-10)*(no. I of bars-1)

=258.66mm

Clear spacing =242.5-half of Dia. -half of Dia bar

=242.5-10-10

Clear spacing =238mm

As per is code value our value is greater so obtained clear spacing is greater than is code value

In order to reduce the value minimize the bar size

Take 16mm bar=4885/20.96

No. of bars=24.30

=24

c/c spacing =(4000-50-50-8-8)(no. I of bars-1)

= 168.89

Clear spacing =168.86-8-8

=152.86mm

So, clear spacing 152.86<180

Hence ok

Step 7-

Check for one way shear

One way shear and the critical section for one way shear at the distance are d from face of the column

Vu=upward pressure x area of the highlight section =206.25 x 4 x ?

Area of highlighted Section=(4000/2)-(400/2)-778

= 1.022

Now, Vu=206.25 x 4 x 1.022

Vu=843.15kN

Nominal stress = Vu/bd

=(843.15 x 10^3)(4000x778)

=0.27N/m

Now, compare Tv and Tc

For the value given by is 456 ,table 19

Pt provided =(100 Ast/bd)

Ast=24 no. Of 16mm dia

=4823mm2

Pt=(100x 4823)/(4000x778)

=0.1549

From is 456 table 19

100A_(st)/bd<0.15is0.28N/mm2

So, 0.15<0.28

Hence ok

Step 8-

check for two way shear at peripheral d/2 distance from the face of the column

area of highlighted section=400+778/2+778/2

=1178mm2

So, area=4^2-1.178^2

= 14.6123mm2

Vu=upwaord pressure x area of highlighted section

=206.25x 14.6123

=3013.89

Tu=Vu/bd

=3013.89/4712*778

=0.8221 N/mm2

Compare Tu and Tc

For two way shear = Ks x Tc

For Tc value taken from IS 456 page no. 58

Ks=1 ....for square column

Tc =0.25√Fck

=1.25 N/mm2

So, Tc=Ks x Tc

=1*1.25

=1.25 N/mm2

So, Tc>Tu

0.8221<1.25

Hence ok