Design of Shallow Foundation (Isolated Footings)

AIM:

To know the general design procedure of spread footing with manual calculations

To understand how the inputs can be colloborated and arrived as data for calculations

INTRODUCTION:

The spread footing behaves like an inverted cantilever with loads applied in the upward direc-tion. As a rule, a spread footing is a quite rigid element therefore, the applied soil stresses are almost linear and in case of a symmetric (with respect to the pedestal) footing, they are or-thogonal. These soil pressures constitutes the loads carried by the footing which behaves like a slab and is deformed.Soil pressures and deformation of an isolated spread footing.

PROCEDURE:

Given:

Column size = 400 x 400 mm

SBC = 150mm^2

Compression axial load=2000KN

`f_(ck)`=M25

Steps to design a square footing:

1. Size of footing

2. Net upward pressure

3. Bending moment

4. Depth of footing

5. Reinforcement

6. One way and two way shear

PROCEDURE:

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`14.66 =3.82

->4mx4m

Therefore provided area=16m2

STEP 2:

Net upward pressure:

`sigma`= Factored load/Area

=1.5xTotal load on footing/area

=1.5x2200/4x4

=206.25kN/m2

STEP 3:

Bending Moment:

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

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

Therefore load on critical section = `sigma` x breadth of footing

=206.25x4

Load=825kN/m

So,the bending moment=wl^2/2

=(825x1.8^2)/2

=1336.5kN-m

STEP 4:

Depth of footing

`M_(u)`=0.138`f_(ck)bd^2`

=0.138x25x4000xd^2

1336.5x10^6= 0.138x25x4000xd^2

d=311.20mm

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

Therefore,

d=778mm

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

=778+50+20+10(Assuming bar dia as 20mm)

D= 860mm

STEP 5:

Reinforcement:

Bending moment = 1336.5 x10^6

Percentage of reinforcement,

`P_(t) = 50 (f_(ck)/f_(y))(1-root'4.6`x13336.5x10^6/25x4000x778^2)

Therefore

`P_(t) = 0.1570%`

Area of steel =`(P_(t)/100)astbd`

=(0.157/100)x4000x778

`A_(st)` = 4885mm^2

No of bars

Take 20mm bar

No of bar = 4885/314.5

=15.55=16 no

Provide 16 nos of 20mm bar

`A_(st)` provided= 16x314.5

=5032mm^2

Therefore

`A_(st)>A_(st)` provided

4885>5032

Hence OK

STEP 6:

Check for clear spacing between bars

As per IS 456-table 15, for Fe 415, the clear spacing between the bars should be less than 180 mm

c/c spacing = (4000-50-50-10-10)/(no. of bars-1)

=258.66mm

Therefore,

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 the min value reduced the bar size

Take 16mm bar=4885/200.96

Therefore

No of bars = 24.3

=24

c/c spacing = (4000-50-50-8-8)/(No 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 at the critical section for one way shear at the distance d from face of the column

`V_(u)`=Upward pressure x Area of highlighted section

=206.25x4xArea of highlighted section

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

=1.022

Now,

`V_(u)` =206.25x4x1.022

`V_(u)`=843.15kN

Nominal stress:

`tau_(v)`= `V_(u)`/bd

=(843.15x10^3)/(4000x778)

`tau_(v)`=0.27 N/m

Now compare

`tau_(v)` and `tau_(c)`

For the `tau_(c)`value given by IS 456,Table 19

`P_(t)`provided =(100 `A_(st)`/bd)

`A_(st)`=24 no of 16mm dia

=4823mm^2

Therefore,

`P_(t)`=(100x4823)/(4000x778)

=0.1549

From table 19 of IS 456

(100 `A_(st)`/bd) value should be `le`0.28N/mm^2

`tau_(v)` <`tau_(c)`

=0.15<0.28

Hence OK

STEP 8:

Check for two way shear

At peripheral d/2 distance from face of the column

Area of highlighted section = 400+(778/2)+(778/2)= 1178 mm^2

Therefore area = 4^2 - 1.178^2

=14.6123mm^2

`V_(u)`=Upward pressure x area of highlighted section

=206.25x14.6123

`V_(u)`=3013.79kN

`tau_(v)=V_(u)/`bd

Peripheral length=4x1178

b=4712mm

`tau_(v)`=(3013.79x10^3)/(4712x778)

`tau_(v)`=0.8221N/mm^2

Compare,

`tau_(v)` and `tau_(v)`

For two way shear

`tau_(c)`=`K_(s) ast tau_(c)`

For `tau_(c)` value taken from IS 456-2000 page number 58

`K_(s)`=1(for square column)

`tau_(c)`=0.25 `root``f_(ck)`

=1.25N/mm^2

Therefore

`tau_(c)`=`K_(s) ast tau_(c)`

=1x1.25

`tau_(c)`=1.25N/mm^2

Therefore

`tau_(v)`<`tau_(c)`

0.8221<1.25

Hence OK

Reinforcement detailing drawing